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79 lines
1.4 KiB
79 lines
1.4 KiB
#include <iostream>
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/*
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测试用例 13 13
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#############
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#GG.GGG#GGG.#
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###.#G#G#G#G#
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#.......#..G#
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#G#.###.#G#G#
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#GG.GGG.#.GG#
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#G#.#G#.#.###
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##G...G.....#
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#G#.#G###.#G#
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#...G#GGG.GG#
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#G#.#G#G#.#G#
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#GG.GGG#G.GG#
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#############
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*/
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using namespace std;
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int main() {
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char a[20][21]; //假设地图大小不超过20*20
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int i,j,sum,map=0,p,q,x,y,n,m;
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cin>>n>>m; //输入行数和列数及各行字符
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for(i=0; i<n; i++)
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for(j=0; j<m; j++)
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cin>>a[i][j];
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for(i=0; i<n; i++) { //两重循环枚举地图每一点
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for(j=0; j<m; j++) {
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if(a[i][j]=='.') { //判断是否为平地
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sum=0; //计数可以消灭的敌人数
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//向上统计可以消灭的敌人数
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x=i;
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y=j;
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while(a[x][y]!='#') { //判断是否为墙，不为墙则继续
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if(a[x][y]=='G') //当前是敌人，则计数
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sum++;
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x--; //继续向上计数
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}
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//向下统计可以消灭的敌人数
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x=i;
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y=j;
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while(a[x][y]!='#') {
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if(a[x][y]=='G')
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sum++;
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x++; //继续向下计数
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}
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//向左统计可以消灭的敌人数
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x=i;
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y=j;
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while(a[x][y]!='#') {
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if(a[x][y]=='G')
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sum++;
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y--; //继续向左计数
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}
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//向右统计可以消灭的敌人数
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x=i;
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y=j;
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while(a[x][y]!='#') {
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if(a[x][y]=='G')
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sum++;
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y++; //继续向右计数
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}
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//更新map的值
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if(sum>map) {
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map=sum;
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p=i; //p、q记录坐标值
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q=j;
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}
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}
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}
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}
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cout<<"将炸弹放置在("<<p<<","<<q<<")处，最多可以消灭"<<map<<"个敌人！"<<endl;
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return 0;
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}
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